Factorial Trailing Zeroes | Leetcode

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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Solution: The idea is to count the number of factors of 5 in all of the numbers from 1 to n inclusive. 5, 10, 15, 20 count as one factor of 5 each. 25 counts as two factors of 5. We count the number of 5's because a zero is obtained by multiplying 2 by 5, and since there are surely more factors of 2 than factors of 5, we only need to count the number of 5'.

A good lengthy explanation beyond what I can write:

      http://www.programmerinterview.com/index.php/java-questions/find-trailing-zeros-in-factorial/

public class Solution {
    public int trailingZeroes(int n) {
        int count = 0;
        while (n >= 5) {
            n /= 5;
            count += n;
        }
        return count;
    }
}